Given-that-ax-b-is-a-factor-of-x-2-2x-2-1-and-a-is-a-root-of-x-2-2x-1-0-show-that-the-value-of-b-is-1-or-3-2-2-

Question Number 40510 by KMA last updated on 23/Jul/18

G i v e n t h a t a x + b i s a f a c t o r o f x^{2}

+ 2 x^{2} - 1 a n d - a i s a r o o t o f x^{2} + 2 x

- 1 = 0 s h o w t h a t t h e v a l u e o f b i s

- 1 o r 3 + 2 \sqrt{2 .}

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

p u t x = \frac{- b}{a} i n x^{2} + 2 x - 1 t h e n R = 0

{(\frac{- b}{a})}^{2} + 2 (\frac{- b}{a}) - 1 = 0

b^{2} - 2 a b - a^{2} = 0

b = \frac{- (- 2 a) \pm \sqrt{4 a^{2} + 4 a^{2}}}{2}

b = a \pm . a \sqrt{2}

p u t x = (- a) i n x^{2} + 2 x - 1 = 0

a^{2} - 2 a - 1 = 0

a = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}

b = a + a \sqrt{2} a n d a - a \sqrt{2}

b = 1 + \sqrt{2} + \sqrt{2} (1 + \sqrt{2}) = 3 + 2 \sqrt{2}

b = 1 - \sqrt{2} + \sqrt{2} (1 - \sqrt{2}) = 1 - \sqrt{2} + \sqrt{2} - 2 = - 1